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在C中通过引用传递指向struct的指针

[英]Pass pointer to struct by reference in C

Take in mind the following piece of code: 请记住以下代码:

#include <stdio.h>
#include <stdlib.h>

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A* a)
{
    a->a = 3;
    a->b = 2;
    a->c = 1;
}
int main()
{
    test = malloc(sizeof(A));
    init(test);
    printf("%d\n", test->a);
    return 0;
}

It runs fine! 它运行正常! Now imagine that I want to use the malloc function outside the main itself without returning a pointer to the struct . 现在假设我想在main之外使用malloc函数而不返回指向struct的指针。 I would put malloc inside init and pass test adress. 我会将malloc放在init并通过test地址。 But this doesnt seem to work. 但这似乎不起作用。

#include <stdio.h>
#include <stdlib.h>

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A** a)
{
    *a = malloc(sizeof(A));
    *a->a = 3;
    *a->b = 2;
    *a->c = 1;
}
int main()
{
    init(&test);
    printf("%d\n", test->a);
    return 0;
}

It keeps telling me that int a (or b / c ) is not a member of the struct A when I use the pointer. 它使告诉我,诠释a (或b / c )不是的成员struct A当我使用指针。

You must add parenthesis: 您必须添加括号:

void init(A **a)
{
    *a = malloc(sizeof(A)); // bad you don't verify the return of malloc
    (*a)->a = 3;
    (*a)->b = 2;
    (*a)->c = 1;
}

But it's good practice to do this: 但这样做是很好的做法:

void init(A **a)
{
    A *ret = malloc(sizeof *ret); // we want the size that is referenced by ret
    if (ret != NULL) { // you should check the return of malloc
        ret->a = 3;
        ret->b = 2;
        ret->c = 1;
    }
    *a = ret;
}

You need to write (*a)->a = 3; 你需要写(*a)->a = 3; for reasons of precedence. 出于优先原因。

Your problem is operator precedence. 您的问题是运营商优先级。 The -> operator has higher precedence than the * (dereference) operator, so *a->a is read as if it is *(a->a) . ->运算符的优先级高于* (取消引用)运算符,因此*a->a的读取就像它是*(a->a) Change *a->a to (*a)->a : *a->a更改为(*a)->a

#include <stdio.h>
#include <stdlib.h>

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A** a)
{
    *a = malloc(sizeof(A));
    (*a)->a = 3;
    (*a)->b = 2;
    (*a)->c = 1;
}
int main()
{
    init(&test);
    printf("%d\n", test->a);
    return 0;
}

Even though it's not a direct answer to your question, since we're in the vicinity of initialization I'd like to point out that C11 gives you a nicer syntax to initialize structs: 即使它不是你问题的直接答案,因为我们正处于初始化的附近,我想指出C11为你提供了一个更好的语法来初始化结构:

void init(A **a)
{
    A *ret = malloc(sizeof *ret); // we want the size that is referenced by ret
    if (ret != NULL) { // you should check the return of malloc
        *ret = (A) {3, 2, 1};
        // or
        *ret = (A) { .a = 3, .b = 2, .c = 1 };
    }
    *a = ret;
}

Another advantage is that any uninitialized members are zeroed. 另一个优点是任何未初始化的成员都归零。

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